Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $p = \dfrac{3x - 27}{x - 5} \times \dfrac{3x^2 - 192}{2x^2 - 34x + 144} $
First factor out any common factors. $p = \dfrac{3(x - 9)}{x - 5} \times \dfrac{3(x^2 - 64)}{2(x^2 - 17x + 72)} $ Then factor the quadratic expressions. $p = \dfrac {3(x - 9)} {x - 5} \times \dfrac {3(x - 8)(x + 8)} {2(x - 8)(x - 9)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac {3(x - 9) \times 3(x - 8)(x + 8) } {(x - 5) \times 2(x - 8)(x - 9) } $ $p = \dfrac {9(x - 8)(x + 8)(x - 9)} {2(x - 8)(x - 9)(x - 5)} $ Notice that $(x - 8)$ and $(x - 9)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {9\cancel{(x - 8)}(x + 8)(x - 9)} {2\cancel{(x - 8)}(x - 9)(x - 5)} $ We are dividing by $x - 8$ , so $x - 8 \neq 0$ Therefore, $x \neq 8$ $p = \dfrac {9\cancel{(x - 8)}(x + 8)\cancel{(x - 9)}} {2\cancel{(x - 8)}\cancel{(x - 9)}(x - 5)} $ We are dividing by $x - 9$ , so $x - 9 \neq 0$ Therefore, $x \neq 9$ $p = \dfrac {9(x + 8)} {2(x - 5)} $ $ p = \dfrac{9(x + 8)}{2(x - 5)}; x \neq 8; x \neq 9 $